Quote:
Originally Posted by DanEC
That would be a tough question to answer - it will depend on the car. You would need to total the area of all the caliper piston bores, assume an amount of piston travel to bring the pads against the rotors and compute a volume of fluid from that. Then convert that volume in to piston travel based on the area of the 1 inch piston in the MC.
For instance - the combined piston area of your front brakes is 11.65 sq inches each or 23.3 in sq for both fronts. I don't know what your rear brakes are but for example use a 2-piston caliper with 2 inch pistons. They would be 6.14 in sq per caliper or 12.28 in sq for both. All brakes would total 35.58 in sq piston area.
Assume the pistons travel 1/32 inch on each side to bring the pads into contact with the rotor - that's 1.112 cu inches of brake fluid.
The 1 inch MC piston is .785 sq inches. 1.112 cu in / .785 in sq = 1.417 inches MC piston travel.
Not sure if that 1/32 inch travel is reasonable or not. The MC definitely should not bottom when applying the brakes.
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Hmm, make sense and means I have way too much travel.
I am using GM Metric rear calipers, the type found on Cadallac Eldorado's with built in parking brake. These calipers have 2.380 in pistons.
To state again, when I pump the brakes the pedal gets quite hard, wait a minute and it soft.